Question

The function $f(x) = [x] \cos \left(\frac{2x-1}{2}\right) \pi$, [.] denotes the greatest integer function, is discontinuous at

• A. All x
• B. All integer points
• C. No x
• D. x which is not an integer

Answer 1

Answer: (C)

No x

Answer 2

When x is not an integer, both the functions [x] and $\cos \left(\frac{2x-1}{2}\right) \pi$ are continuous.
$\therefore \, f(x)$ is continuous on all non integral points.
For $x = n \in I$
$\displaystyle \lim_{x \to n^{-}} f\left(x\right) =\displaystyle \lim_{x \to n^{-}} \left[x\right]\cos\left(\frac{2x-1}{2}\right) \pi$
$=\left(n-1\right) \cos \left(\frac{2-1}{2}\right) \pi= 0$
$\displaystyle \lim_{x \to n^{+}} f\left(x\right) =\displaystyle \lim_{x \to n^{+}} \left[x\right]\cos\left(\frac{2x-1}{2}\right) \pi$
$= n \cos\left(\frac{2n-1}{2}\right) \pi= 0$
Also $f\left(n\right) = n \cos \frac{\left(2n-1\right)\pi}{2} = 0$
$\therefore \, f$ is continuous at all integral pts as well.
Thus, $f$ is continuous everywhere.